\section{合同}
	\begin{titwo}
		实二次型 $f(x_{1},x_{2},\cdots,x_{n})$ 的秩为 $r$，符号差为 $s$，且 $f$ 和 $-f$ 对应的矩阵合同，则必有 \kuo.

		\twoch{$r$ 是偶数，$s = 1$}{$r$ 是奇数，$s = 1$}{$r$ 是偶数，$s = 0$}{$r$ 是奇数，$s = 0$}
	\end{titwo}

	\begin{titwo}
		设方阵 $\bm A_{1}$ 与 $\bm B_{1}$ 合同，$\bm A_{2}$ 与 $\bm B_{2}$ 合同，证明：$\begin{bsmallmatrix}
			\bm A_{1} & \\
			& \bm A_{2}
		\end{bsmallmatrix}$ 与 $\begin{bsmallmatrix}
			\bm B_{1} & \\
			& \bm B_{2}
		\end{bsmallmatrix}$ 合同.
	\end{titwo}

	\begin{titwo}
		设 $\bm A$, $\bm B$ 是 $n$ 阶实对称可逆矩阵，则存在 $n$ 阶可逆阵 $\bm P$，使得下列关系式 \circled{1}~$\bm P \bm A = \bm B$; \circled{2}~$\bm P^{-1} \bm A \bm B \bm P = \bm B \bm A$; \circled{3}~$\bm P^{-1} \bm A \bm P = \bm B$; \circled{4}~$\bm P^{\TT} \bm A^{2} \bm P = \bm B^{2}$ 成立的个数是 \kuo.

		\fourch{1}{2}{3}{4}
	\end{titwo}

	\begin{titwo}
		有三组二次型\\
		\circled{1}~$f(x_{1},x_{2},x_{3}) = x_{1}^{2} + 4x_{1}x_{2} + x_{2}^{2} + x_{3}^{2}$, $g(y_{1},y_{2},y_{3}) = y_{1}^{2} + y_{2}^{2} + 2y_{2}y_{3} + y_{3}^{2}$;\\
		\circled{2}~$f(x_{1},x_{2},x_{3}) = \lambda_{1} x_{1}^{2} + \lambda_{2} x_{2}^{2} + \lambda_{3} x_{3}^{2}$, $g(y_{1},y_{2},y_{3}) = \lambda_{3} y_{1}^{2} + \lambda_{1} y_{2}^{2} + \lambda_{2} y_{3}^{2}$;\\
		\circled{3}~$f(x_{1},x_{2},x_{3}) = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}$, $g(y_{1},y_{2},y_{3}) = y_{2}^{2} + 2y_{1}y_{3}$.\\
		二次型矩阵彼此合同的有 \kuo.

		\fourch{0 组}{1 组}{2 组}{3 组}
	\end{titwo}

	\begin{titwo}
		设 $3$ 阶实对称矩阵
		\[
			\bm A = \begin{bsmallmatrix}
				a_{1} + a_{2} + a_{3} & a_{2} + a_{3} & a_{3} \\
				a_{2} + a_{3} & a_{2} + a_{3} & a_{3} \\
				a_{3} & a_{3} & a_{3}
			\end{bsmallmatrix},
			\bm B = \begin{bsmallmatrix}
				k_{3}a_{1} & 0 & 0 \\
				0 & k_{2}a_{2} & 0 \\
				0 & 0 & k_{1}a_{3}
			\end{bsmallmatrix},
		\]
		其中 $k_{1}$, $k_{2}$, $k_{3}$ 为大于 $0$ 的任意常数. 证明 $\bm A$ 与 $\bm B$ 合同，并求出可逆矩阵 $\bm C$，使得 $\bm C^{\TT} \bm A \bm C = \bm B$.
	\end{titwo}